Suppose that \(d\)  is odd, and denote by \( L_{m, d}^*\) , \(H_{m, d}\)  the genus field and the Hilbert 2-class field of \(L_{m, d}\)  respectively. It is known that:
\([L_{m, d}^*:\mathbb {Q}]=\prod _{p|\delta _{L_{m, d}}}e(p)\quad \text{ and }\quad \mathrm {Cl}_2(L_{m, d})=Gal(H_{m, d}/L_{m, d}), \) 
where \(e(p)\)  is the ramification index of \(p\)  in \(L_{m, d}\) . So
\([L_{m, d}^*:\mathbb {Q}]=\prod _{p|2d}e(p)=[L_{m, d}^*:L_{m, d}][L_{m, d}:\mathbb {Q}]=2^{m}[L_{m, d}^*:L_{m, d}].\) 
Since \(e(2)=2^{m-1}\)  and \(e(p)=2\)  for any prime divisor \(p\)  of \(d\) , we have
\(\prod _{p|d}e(p)=2.[L_{m, d}^*:L_{m, d}].\) 
Hence, if \(d\)  is not a prime, then \(L_{m, d}\subsetneq L_{m, d}^*\subseteq H_{m, d}\) 
and \(h_2(L_{m, d})\)  is even.
Suppose now that \(d=p\)  is a prime. We distinguish the following four cases:
\(\bullet \)  Assume \(d=p\equiv 1\pmod {8}\) . Set \(p=a^2+16b^2=e^2-32f^2\)  and \(\pi _1=a+4bi\) , \(\pi _2= e+4f\sqrt{2}\) . As the ramified primes of \(K_m\)  in \(L_{m, d}\)  are exactly the prime divisors of \(p\)  in \(K_m\) , then the ideals of \(L_{m, d}\)  generated by \(\pi _1\)  and \(\pi _2\)  are squares of ideals of \(L_{m, d}\) . Note that as \(a\)  and \(e\)  are odd, then \(a\equiv e \equiv \pm 1\equiv i^2\pmod {4}\) . It follows that the equation \(\pi _j\equiv \xi ^2\pmod {4}\) , \(j=1\)  or 2, has a solution. So \(L_1=L_{m, d}(\sqrt{\pi _1})\)  and \(L_2=L_{m, d}(\sqrt{\pi _2})\)  are two distinct unramified quadratic extensions of \(L_{m, d}\) . Thus \(h(L_{m, d})\)  is divisible by 4. Furthermore, \(\mathrm {Cl}_2(L_{m, d})\)  is not trivial and not cyclic.
\(\bullet \)  Assume now \(d=p\equiv 7\pmod {8}\) .
We prove that \(h(L_{m, p})\)  is even for all \(m\ge 3\)  by induction on \(m\) . If \(m=3\) , then \(h(L_{3, p})\)  is even by [1]}. Suppose that \(h(L_{m, p})\)  is even for some \(m\ge 3\) . We have
\(\mathbb {Q}(\sqrt{-p})/\mathbb {Q}\)  is unramified at 2 and \(\mathbb {Q}(\zeta _{2^{m}})/\mathbb {Q}\)  is totally ramified at 2, then \(L_{m+1, p}/L_{m, p}\)  is a quadratic extension that is ramified at primes over 2. So \(h(L_{m, p})\)  divides \(h(L_{m+1, p})\) , by Theorem REF . Hence, \(h(L_{m+1, p})\)  is even.

[scale=1]
Q) at (0, -0.5) \(\mathbb {Q}\) ;
Q1) at (-0.3, 0.3) \(\mathbb {Q}(\sqrt{-p})\) ;
Q2) at (1.5, 0.5) \(\mathbb {Q}(\zeta _{2^{m}})\) ;
Q4) at (0.5, 1.5) \(\mathbb {Q}(\sqrt{p}, \zeta _{2^{m}})\) ;
Q5) at (2, 1.5) \(\mathbb {Q}(\zeta _{2^{m+1}})\) ;
Q7) at (2, 2.5) \(\mathbb {Q}(\sqrt{p}, \zeta _{2^{m+1}})\) ;
(Q) –(Q1) node[scale=0.4, midway, below right];
(Q) –(Q2) node[scale=0.4, midway, below right];
(Q2) –(Q4) node[scale=0.4, midway, below right];
(Q4) –(Q7) node[scale=0.4, midway, below right];
(Q2) –(Q5) node[scale=0.4, midway, below right];
(Q5) –(Q7) node[scale=0.4, midway, below right];
(Q1) –(Q4) node[scale=0.4, midway, below right];

\(\bullet \)  Assume that \(d=p\equiv 5\pmod {8}\) . For \(m\ge 3\) , we have \(p\)  decomposes into the product of two prime ideals of \(K_m\) , denote by \(\mathfrak {p}_{K_m}\)  one of them (such that \(\mathfrak {p}_{K_{m-1}}\subset \mathfrak {p}_{K_m}\) ). Since \(\zeta _{2^{m}}^2=\zeta _{2^{{m-1}}}\) , so the minimal polynomial of \(\zeta _{2^{m}}\)  over \(K_{{m-1}}\)  is \(X^2-\zeta _{2^{m-1}}\)  and \(N_{K_{m}/K_{m-1}}(\zeta _{2^{m}})=-\zeta _{2^{m-1}}\) .
Then
\(\left( \frac{\zeta _{2^{m}}, p}{\mathfrak {p}_{K_m}}\right)=\left( \frac{-\zeta _{ 2^{m-1}}, p}{\mathfrak {p}_{K_{m-1}}}\right)=\left( \frac{\zeta _{2^{m-1}}, p}{\mathfrak {p}_{K_{m-1}}}\right)=...=\left( \frac{\zeta _{8}, p}{\mathfrak {p}_{K_3}}\right)=-1, \) 
hence \(e_{m, d}\ne 0\)  and \(rank_2(Cl(L_{m+1, p}))=2-1-e_{m+1, p}=1-e_{m+1, p} =0\) . Thus the 2-class group of \(L_{m+1, p}\)  is trivial and \(h(L_{m, d})\)  is odd.
\(\bullet \)  We treat the case \(d=p\equiv 3\pmod {8}\) , similarly to the previous one and we show that \(h(L_{m, d})\)  is odd. Which achieves the proof.
